An atom (or ion) has a ground state energy of -54.4 eV . This species could be

Correct answer: D. He^+ The energy of an electron in the n^ th orbit of a hydrogen-like species is given by E_n = -13.6 (Z^2)/(n^2) eV . For the ground state, n = 1 . E_1 = -13.6 Z^2 eV Given E_1 = -54.4 eV . -13.6 Z^2 = -54.4 Z^2 = (54.4)/(13.6) = 4 Z = 2 The species with atomic number Z = 2 is helium. Since the Bohr model applies to single-electron species, the ion is He^+ .

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Bohr's Model and Physics of the Atom

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Bohr's Model and Physics of the Atom

An atom (or ion) has a ground state energy of -54.4 eV . This...

An atom (or ion) has a ground state energy of

-54.4 \text{ eV}

. This species could be

Options

Adeuterium
B $\text{Li}^{++}$
Chydrogen
D $\text{He}^+$

Correct answer

D. $\text{He}^+$

Step-by-step solution

The energy of an electron in the

n^{\text{th}}

orbit of a hydrogen-like species is given by

E_n = -13.6 \dfrac{Z^2}{n^2} \text{ eV}

.

For the ground state,

n = 1

E_1 = -13.6 Z^2 \text{ eV}

Given

E_1 = -54.4 \text{ eV}

-13.6 Z^2 = -54.4

Z^2 = \dfrac{54.4}{13.6} = 4

Z = 2

The species with atomic number

Z = 2

is helium. Since the Bohr model applies to single-electron species, the ion is

\text{He}^+

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